Problem: The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $10.4$ years; the standard deviation is $1.9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living longer than $16.1$ years.
$10.4$ $8.5$ $12.3$ $6.6$ $14.2$ $4.7$ $16.1$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $10.4$ years. We know the standard deviation is $1.9$ years, so one standard deviation below the mean is $8.5$ years and one standard deviation above the mean is $12.3$ years. Two standard deviations below the mean is $6.6$ years and two standard deviations above the mean is $14.2$ years. Three standard deviations below the mean is $4.7$ years and three standard deviations above the mean is $16.1$ years. We are interested in the probability of a meerkat living longer than $16.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the meerkats will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the meerkats will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $4.7$ years and the other half $({0.15\%})$ will live longer than $16.1$ years. The probability of a particular meerkat living longer than $16.1$ years is ${0.15\%}$.